RS Aggarwal (2021 2022) Solutions for Class 9 Maths Chapter 8 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among class 9 students for Maths Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal (2021 2022) Book of class 9 Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal (2021 2022) Solutions. All RS Aggarwal (2021 2022) Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 252:
Question 1:
Answer:
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Question 2:
Answer:
Let the angles of the given triangle measure , respectively.
Then,
Hence, the measures of the angles are .
Page No 252:
Question 3:
Let the angles of the given triangle measure , respectively.
Then,
Hence, the measures of the angles are .
Answer:
Let .
Then,
Therefore,
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Question 4:
Let .
Then,
Therefore,
Answer:
Let .
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Question 5:
Let .
Answer:
Let .
Then,
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Question 6:
Let .
Then,
Answer:
Then,
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Question 7:
Then,
Answer:
Let
Then,
Also, in â ABC:
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Question 8:
Let
Then,
Also, in â ABC:
Answer:
Let .
Then,
Since,
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Question 9:
Let .
Then,
Since,
Answer:
Let the smallest angle of the triangle be and let .
Then,
Also,
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Question 10:
Let the smallest angle of the triangle be and let .
Then,
Also,
Answer:
Let ABC be a triangle right-angled at B.
Then,
Hence, .
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Question 11:
Let ABC be a triangle right-angled at B.
Then,
Hence, .
Answer:
Let ABC be a triangle.
Then,
This implies that the triangle is right-angled at A.
Page No 252:
Question 12:
Let ABC be a triangle.
Then,
This implies that the triangle is right-angled at A.
Answer:
Let ABC be the triangle.
Let
Then,
Also, let
Then,
And let
Then,
Hence, each angle of the triangle is less than .
Therefore, the triangle is acute-angled.
Page No 252:
Question 13:
Let ABC be the triangle.
Let
Then,
Also, let
Then,
And let
Then,
Hence, each angle of the triangle is less than .
Therefore, the triangle is acute-angled.
Answer:
Let ABC be a triangle and let .
Then, we have:
Since one of the angles of the triangle is greater than , the triangle is obtuse-angled.
Page No 252:
Question 14:
Let ABC be a triangle and let .
Then, we have:
Since one of the angles of the triangle is greater than , the triangle is obtuse-angled.
Answer:
Side BC of triangle ABC is produced to D.
Also, in triangle ABC,
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Question 15:
Side BC of triangle ABC is produced to D.
Also, in triangle ABC,
Answer:
Side BC of triangle ABC is produced to D.
Also, side BC of triangle ABC is produced to E.
And,
Page No 253:
Question 16:
Side BC of triangle ABC is produced to D.
Also, side BC of triangle ABC is produced to E.
And,
Answer:
(i)
Side AC of triangle ABC is produced to E.
Also,
Substituting the value of
(ii)
From we have:
Also,
(iii)
Also,
(iv)
(v)
From , we have:
Also, from , we have:
(vi)
From , we have:
Also, From , we have
Page No 253:
Question 17:
(i)
Side AC of triangle ABC is produced to E.
Also,
Substituting the value of
(ii)
From we have:
Also,
(iii)
Also,
(iv)
(v)
From , we have:
Also, from , we have:
(vi)
From , we have:
Also, From , we have
Answer:
In the given figure, AB || CD and AC is the transversal.
∴ ∠ACD = ∠BAC = 60º (Pair of alternate angles)
Or ∠GCH = 60º
Now, ∠GHC = ∠DHF = 50º (Vertically opposite angles)
In âGCH,
∠AGH = ∠GCH + ∠GHC (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)
⇒ ∠AGH = 60º + 50º = 110º
Page No 253:
Question 18:
In the given figure, AB || CD and AC is the transversal.
∴ ∠ACD = ∠BAC = 60º (Pair of alternate angles)
Or ∠GCH = 60º
Now, ∠GHC = ∠DHF = 50º (Vertically opposite angles)
In âGCH,
∠AGH = ∠GCH + ∠GHC (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)
⇒ ∠AGH = 60º + 50º = 110º
Answer:
![](/img/site_content/ask-answer/loader.gif)
Join A and D to produce AD to E.
Then,
Side AD of triangle ACD is produced to E.
(Exterior angle property)
Side AD of triangle ABD is produced to E.
(Exterior angle property)
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Question 19:
![](/img/site_content/ask-answer/loader.gif)
Join A and D to produce AD to E.
Then,
Side AD of triangle ACD is produced to E.
(Exterior angle property)
Side AD of triangle ABD is produced to E.
(Exterior angle property)
Answer:
Now, divide in the ratio 1 : 3.
Hence, the angles are 18o and 54o
Given,
In , we have:
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Question 20:
Now, divide in the ratio 1 : 3.
Hence, the angles are 18o and 54o
Given,
In , we have:
Answer:
Side BC of triangle ABC is produced to D.
Side AC of triangle ABC is produced to E.
And side AB of triangle ABC is produced to F.
Hence, the sum of the exterior angles so formed is equal to four right angles.
Page No 254:
Question 21:
Side BC of triangle ABC is produced to D.
Side AC of triangle ABC is produced to E.
And side AB of triangle ABC is produced to F.
Hence, the sum of the exterior angles so formed is equal to four right angles.
Answer:
In , we have :
[Sum of the angles of a triangle]
In , we have :
[Sum of the angles of a triangle]â
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Question 22:
In , we have :
[Sum of the angles of a triangle]
In , we have :
[Sum of the angles of a triangle]â
Answer:
In , we have:
In , we have:
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Question 23:
In , we have:
In , we have:
Answer:
In the given figure, EF || BD and CE is the transversal.
∴ ∠CAD = ∠AEF (Pair of corresponding angles)
⇒ ∠CAD = 55°
In âABC,
∠CAD = ∠ABC + ∠ACB (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)
⇒ 55° = ∠ABC + 25°
⇒ ∠ABC = 55° − 25° = 30°
Thus, the measure of ∠ABC is 30°.
Page No 255:
Question 24:
In the given figure, EF || BD and CE is the transversal.
∴ ∠CAD = ∠AEF (Pair of corresponding angles)
⇒ ∠CAD = 55°
In âABC,
∠CAD = ∠ABC + ∠ACB (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)
⇒ 55° = ∠ABC + 25°
⇒ ∠ABC = 55° − 25° = 30°
Thus, the measure of ∠ABC is 30°.
Answer:
Let
From , we have:
Side BC of triangle ABC is produced to E.
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Question 25:
Let
From , we have:
Side BC of triangle ABC is produced to E.
Answer:
We have, ∠FGE + ∠FGH = 180° (Linear pair of angles)
∴ y + 120° = 180°
⇒ y = 180° − 120° = 60°
Now, AB || DF and BD is the transversal.
∴ ∠ABD = ∠BDF (Pair of alternate angles)
⇒ ∠BDF = 50°
Also, BD || FG and DF is the transversal.
∴ ∠BDF = ∠DFG (Pair of alternate angles)
⇒ ∠DFG = 50° .....(1)
In âEFG,
∠FGH = ∠EFG + ∠FEG (Exterior angle of a triangle is equal to the sum of the two opposite interior angles)
⇒ 120° = 50° + x [Using (1)]
⇒ x = 120° − 50° = 70°
Thus, the values of x and y are 70° and 60°, respectively.
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Question 26:
We have, ∠FGE + ∠FGH = 180° (Linear pair of angles)
∴ y + 120° = 180°
⇒ y = 180° − 120° = 60°
Now, AB || DF and BD is the transversal.
∴ ∠ABD = ∠BDF (Pair of alternate angles)
⇒ ∠BDF = 50°
Also, BD || FG and DF is the transversal.
∴ ∠BDF = ∠DFG (Pair of alternate angles)
⇒ ∠DFG = 50° .....(1)
In âEFG,
∠FGH = ∠EFG + ∠FEG (Exterior angle of a triangle is equal to the sum of the two opposite interior angles)
⇒ 120° = 50° + x [Using (1)]
⇒ x = 120° − 50° = 70°
Thus, the values of x and y are 70° and 60°, respectively.
Answer:
It is given that, AB || CD and EF is a transversal.
∴ ∠EFD = ∠AEF (Pair of alternate angles)
⇒ ∠EFD = 65°
⇒ ∠EFG + ∠GFD = 65°
⇒ ∠EFG + 30° = 65°
⇒ ∠EFG = 65° − 30° = 35°
In âEFG,
∠EFG + ∠GEF + ∠EGF = 180° (Angle sum property)
⇒ 35° + x + 90° = 180°
⇒ 125° + x = 180°
⇒ x = 180° − 125° = 55°
Thus, the value of x is 55°.
Page No 255:
Question 27:
It is given that, AB || CD and EF is a transversal.
∴ ∠EFD = ∠AEF (Pair of alternate angles)
⇒ ∠EFD = 65°
⇒ ∠EFG + ∠GFD = 65°
⇒ ∠EFG + 30° = 65°
⇒ ∠EFG = 65° − 30° = 35°
In âEFG,
∠EFG + ∠GEF + ∠EGF = 180° (Angle sum property)
⇒ 35° + x + 90° = 180°
⇒ 125° + x = 180°
⇒ x = 180° − 125° = 55°
Thus, the value of x is 55°.
Answer:
In the given figure, AB || CD and AE is the transversal.
∴ ∠DOE = ∠BAE (Pair of corresponding angles)
⇒ ∠DOE = 65°
In âCOE,
∠DOE = ∠OEC + ∠ECO (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
⇒ 65° = 20° + ∠ECO
⇒ ∠ECO = 65° − 20° = 45°
Thus, the measure of ∠ECO is 45°.
Page No 255:
Question 28:
In the given figure, AB || CD and AE is the transversal.
∴ ∠DOE = ∠BAE (Pair of corresponding angles)
⇒ ∠DOE = 65°
In âCOE,
∠DOE = ∠OEC + ∠ECO (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
⇒ 65° = 20° + ∠ECO
⇒ ∠ECO = 65° − 20° = 45°
Thus, the measure of ∠ECO is 45°.
Answer:
In the given figure, AB || CD and EF is a transversal.
∴ ∠PHQ = ∠EGB (Pair of alternate exterior angles)
⇒ ∠PHQ = 35°
In âPHQ,
∠PHQ + ∠QPH + ∠PQH = 180° (Angle sum property)
⇒ 35° + 90° + x = 180°
⇒ 125° + x = 180°
⇒ x = 180° − 125° = 55°
Thus, the measure of ∠PQH is 55°.
Page No 255:
Question 29:
In the given figure, AB || CD and EF is a transversal.
∴ ∠PHQ = ∠EGB (Pair of alternate exterior angles)
⇒ ∠PHQ = 35°
In âPHQ,
∠PHQ + ∠QPH + ∠PQH = 180° (Angle sum property)
⇒ 35° + 90° + x = 180°
⇒ 125° + x = 180°
⇒ x = 180° − 125° = 55°
Thus, the measure of ∠PQH is 55°.
Answer:
In the given figure, AB || CD and GE is the transversal.
∴ ∠GED + ∠EGF = 180° (Sum of adjacent interior angles on the same side of the transversal is supplementary)
⇒ 130° + ∠EGF = 180°
⇒ ∠EGF = 180° − 130° = 50°
Thus, the measure of ∠EGF is 50°.
Page No 257:
Question 1:
In the given figure, AB || CD and GE is the transversal.
∴ ∠GED + ∠EGF = 180° (Sum of adjacent interior angles on the same side of the transversal is supplementary)
⇒ 130° + ∠EGF = 180°
⇒ ∠EGF = 180° − 130° = 50°
Thus, the measure of ∠EGF is 50°.
Answer:
LCM of 3, 4 and 6 = 12
3∠A = 4∠B = 6∠C (Given)
Dividing throughout by 12, we get
Let , where k is some constant
Then, ∠A = 4k, ∠B = 3k, ∠C = 2k
∴ ∠A : ∠B : ∠C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option (b).
Page No 258:
Question 2:
LCM of 3, 4 and 6 = 12
3∠A = 4∠B = 6∠C (Given)
Dividing throughout by 12, we get
Let , where k is some constant
Then, ∠A = 4k, ∠B = 3k, ∠C = 2k
∴ ∠A : ∠B : ∠C = 4k : 3k : 2k = 4 : 3 : 2
Hence, the correct answer is option (b).
Answer:
(c) 53°
Let
Page No 258:
Question 3:
(c) 53°
Let
Answer:
Hence, the correct answer is option (b).
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Question 4:
Hence, the correct answer is option (b).
Answer:
(d) 75°
We have :
Also,
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Question 5:
(d) 75°
We have :
Also,
Answer:
(a) 65°
We have :
Side AB of triangle ABC is produced to E.
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Question 6:
(a) 65°
We have :
Side AB of triangle ABC is produced to E.
Answer:
(d) 360°
We have :
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Question 7:
(d) 360°
We have :
Answer:
In the given figure, ∠CAD = ∠EAF (Vertically opposite angles)
∴ ∠CAD = 30°
In âABD,
∠ABD + ∠BAD + ∠ADB = 180° (Angle sum property)
⇒ (x + 10)° + (x° + 30°) + 90° = 180°
⇒ 2x° + 130° = 180°
⇒ 2x° = 180° − 130° = 50°
⇒ x = 25
Thus, the value of x is 25.
Hence, the correct answer is option (b).
Page No 259:
Question 8:
In the given figure, ∠CAD = ∠EAF (Vertically opposite angles)
∴ ∠CAD = 30°
In âABD,
∠ABD + ∠BAD + ∠ADB = 180° (Angle sum property)
⇒ (x + 10)° + (x° + 30°) + 90° = 180°
⇒ 2x° + 130° = 180°
⇒ 2x° = 180° − 130° = 50°
⇒ x = 25
Thus, the value of x is 25.
Hence, the correct answer is option (b).
Answer:
In the given figure, ∠ABF + ∠ABC = 180° (Linear pair of angles)
∴ x° + ∠ABC = 180°
⇒ ∠ABC = 180° − x° .....(1)
Also, ∠ACG + ∠ACB = 180° (Linear pair of angles)
∴ y° + ∠ACB = 180°
⇒ ∠ACB = 180° − y° .....(2)
Also, ∠BAC = ∠DAE = z° .....(3) (Vertically opposite angles)
In âABC,
∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property)
∴ z° + 180° − x° + 180° − y° = 180° [Using (1), (2) and (3)]
⇒ z = x + y − 180
Hence, the correct answer is option (a).
Page No 259:
Question 9:
In the given figure, ∠ABF + ∠ABC = 180° (Linear pair of angles)
∴ x° + ∠ABC = 180°
⇒ ∠ABC = 180° − x° .....(1)
Also, ∠ACG + ∠ACB = 180° (Linear pair of angles)
∴ y° + ∠ACB = 180°
⇒ ∠ACB = 180° − y° .....(2)
Also, ∠BAC = ∠DAE = z° .....(3) (Vertically opposite angles)
In âABC,
∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property)
∴ z° + 180° − x° + 180° − y° = 180° [Using (1), (2) and (3)]
⇒ z = x + y − 180
Hence, the correct answer is option (a).
Answer:
In the given figure, ∠BOD = ∠COA (Vertically opposite angles)
∴ ∠BOD = 40° .....(1)
In âACO,
∠OAE = ∠OCA + ∠COA (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
⇒ x° = 80° + 40° = 120° .....(2)
In âBDO,
∠DBF = ∠BDO + ∠BOD (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
⇒ y° = 70° + 40° = 110° [Using (1)] .....(3)
Adding (2) and (3), we get
x° + y° = 120° + 110° = 230°
Hence, the correct answer is option (b).
Page No 259:
Question 10:
In the given figure, ∠BOD = ∠COA (Vertically opposite angles)
∴ ∠BOD = 40° .....(1)
In âACO,
∠OAE = ∠OCA + ∠COA (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
⇒ x° = 80° + 40° = 120° .....(2)
In âBDO,
∠DBF = ∠BDO + ∠BOD (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
⇒ y° = 70° + 40° = 110° [Using (1)] .....(3)
Adding (2) and (3), we get
x° + y° = 120° + 110° = 230°
Hence, the correct answer is option (b).
Answer:
Let
Then,
Hence, the angles are
Side BC of triangle ABC is produced to E.
Hence, the correct answer is option (a).
Page No 259:
Question 11:
Let
Then,
Hence, the angles are
Side BC of triangle ABC is produced to E.
Hence, the correct answer is option (a).
Answer:
(c) 115°
In , we have:
In , we have:
Page No 260:
Question 12:
(c) 115°
In , we have:
In , we have:
Answer:
Disclaimer: In the question ∠ACD should be 7y°.
In the given figure, ∠ACB + ∠ACD = 180° (Linear pair of angles)
∴ 5y° + 7y° = 180°
⇒ 12y° = 180°
⇒ y = 15 .....(1)
In âABC,
∠A + ∠B + ∠ACB = 180° (Angle sum property)
∴ 3y° + x° + 5y° = 180°
⇒ x° + 8y° = 180°
⇒ x° + 8 × 15° = 180° [Using (1)]
⇒ x° + 120° = 180°
⇒ x° = 180° − 120° = 60°
Thus, the value of x is 60.
Hence, the correct answer is option (a).
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